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A particle of mass m kg and charge q coulomb is accelerated through V volt then the de-Broglie wavelength associated with it,in meter is lambda=…..m. |
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Answer» <P>`lambda=(h)/(mv)` `(p^(2))/(2M)=qVimpliesp=sqrt(2Vqm)`……..(2) PUTTING vallue of EQUATION (2) in equation (1), `therefore lambda=(h)/(sqrt(2Vqm))` |
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